Download Probability Methods in Civil Engineering Prof. Dr. Rajib Maity

Probability Methods in Civil Engineering
Prof. Dr. Rajib Maity
Department of Civil Engineering
Indian Institution of Technology, Kharagpur
Lecture No. # 31
Probability Models using Normal Distribution
Welcome to this course on probability methods in civil engineering. Today, we are
starting a new module and this module is on common probability models and in this
module, we will take whatever the standard distributions that we discuss in module 4.
We will take those models, those distributions and we will show some practical
application in different fields of civil engineering. So, today that this lecture mainly we
will be talking about probability models using normal distribution and this distribution
when we are taking the normal distribution, you know that its different, it is different
properties and so far as this is p d f and c d f is concerned, we have discussed in earlier
module and in this lecture also, we will just briefly see those things and mainly our focus
for today’s lecture will be show some practical civil engineering related problem, in
which this properties of this normal distribution has been used. That we will see.
(Refer Slide Time: 00:57)
Basically in this module, we will take up both the discrete random variable as well as
continuous random variable, both and their standard distribution, that we discussed
earlier. So, we are starting with the normal distribution, because this is so far as the
concept of that continuous random variable, will be easier for to start with this normal
distribution.
(Refer Slide Time: 01:58)
So, our the outline for today’s lecture will be, we will first briefly discuss about this
normal distribution and how we calculate the probability because you can recall from our
earlier lecture, that for this normal distribution, the calculation of probability or the
integration, the integration form is not is not available in the closed forms. So, we have
to go for some numerical integration and there are some standard tables that we
introduced. So, those tables have to be used. So, basically after that, we will discuss
some of the problems related to the civil engineering, gradually we will also see the
other, which are some of the very important theorem in the context of this normal
distribution for example, the central limit theorem. That theorem will discuss, may be not
in this lecture, but the subsequent lectures of this module.
And after that, we will take up this probability plotting, basically when we see that
probability plotting, that in that probability plotting, what we do? We generally test. This
is not specific to the normal distribution, this can be for any distribution and probability
plotting means, we generally plot the probability on a special paper just to investigate,
whether the data that we are handling with, whether that is following a particular
distribution or not. Basically, in this in this lecture, while discussing those problems, we
generally assume that data set is following a normal distribution. So, that means, the
problems are started with the assumption of this normal distribution.
So, the question that remain unanswered is that how do you test that whether really the
data that we are talking about is following a particular distribution, in this case, the
normal distributions. So, that is generally tested through different statistical test and
those test, we will take up in this next module and in this one, what we will do, we will
show some the concept of this probability plotting and so, through which, we can test
that from which distribution the data set is drawn. So, here in the context of this normal
distribution, we will we will focus on this normal probability plot, but this will be taken
in the subsequent lectures. Today, we will specifically concern about that normal
distribution at their some of the specific application in the different field of civil
engineering.
(Refer Slide Time: 04:36)
So, as you know from the earlier lecture, that this normal distribution is having two
parameters, one you know, that is the mu and another parameter is called the variances.
So, using those two parameters, this is a bell shaped curve and the p d f of that normal
distribution is from this one, where these two parameters are there mu and sigma square.
So, and this is a continuous probability distribution you know that and this continuous
probability, density function with two parameters, one is that mean, which is denoted by
mu and other one is variance denoted by which is denoted by this sigma square.
And now, this probability density function is given by in this form, that is f x equals to f
x mu is equals to 1 by square root of 2 pi sigma, sigma is outside this root and then e
power minus half x minus mu by sigma whole square. And this is a support of this
distribution which is from minus infinity to plus infinity. And, if we want to know that
that cumulative distribution function, which is c d f at any specific for a specific value,
which is small x, you know that this is a integration from this minus infinity to that
specific point and this integral from. And this is the integration which, for which that any
close from solution or the analytical solution is not available and we have also discussed
that, there are some different, that the numerical integration is done and that is presented
in terms of some tables, that table also we will see in a minute.
(Refer Slide Time: 06:15)
Before that, these are you know, that this is a typical p d f of a normal distribution, how
the p d f looks like, this is a this is bell shape curve and it is symmetrical with respect to
its mean and its typical c d f that is distribution, that cumulative distribution function,
which is starts from 0 at minus infinity and go gradually increases and it is asymptotic to
1 at plus infinity.
(Refer Slide Time: 06:58)
So, now the p d f of the normal distribution is a bell shaped curve, that you have seen,
that it is symmetric about its mean and the mean mu is the location parameter of the
distribution and the change in the values of the mean results in the shift of this normal p
d f curve along the x axis. So, that what happens is that? So, here you can see that may
be the mean is around. So, mean is at 1. So, if we just change it. So, at this mean, this is
the maximum, you can say that the value of this probability density function is maximum
at this means, so if we do not change its sigma. So, this over all shape of this curve will
not change, only that that depending on what the mean is this curve will be, will be
relocated or this will be shifted.
So, that is why, this mu is known as your, that location parameter and whereas, that this
variance, that is sigma square, this sigma square is the scale parameter, why this scale
parameter, because it is generally it controls that how what is the shape of that
distribution and if we change this parameter, that is if the variance is changed, then the
that it will result in the change in the spread of the normal p d f curve. So, the change in
the spread of this normal p d f curve means. It will be, so, if the mean is same. So, its
maximum value of this p d f will be same at this location only, but its spread will be
changed. So, you have seen it earlier. So, the curve will look like this.
(Refer Slide Time: 08:46)
So, if this is your x axis, then if the 1 normal distribution looks like this and if this its
mean and if the scale parameter changes, then the then depending on whether it is
increasing or decreasing, this will be either, it will be flatter or it may be even the even
be more, the spread will be much lesser. So, this is for the minimum, among these 3
curves, this is for the minimum variance and this is for the medium and this is the
maximum variance is here, but all, for all these the mean is same, the location parameter
is same. So, this is what is explained here, that variance sigma square is the scale
parameter of this distribution and change in values of the variance, will result in a change
in the spread of the normal p d f curve.
(Refer Slide Time: 09:51)
Now, as we have shown earlier also, that for this random variable x, which follows a
normal distribution with parameter mean is mu and the and other one is sigma square,
then the probability of x being less than or equal to a particular value x is given by the
probability x less than equals to that particular value, which is the integration from this
left extreme, that is minus infinity to x and from 1 by this form, that p d f form, 1 by
square root pi sigma e power minus half x minus mu by sigma whole square d x. So, this
integration, if we do it for any specific value, putting any specific value x, then we will
get its c d f, that is cumulative distribution function.
Now, this equation, for this normal probability density, cannot be integrated analytically
as I told in the in the close form. Thus the value of the probability is to be computed by
numerical integration. So, once this mu, these parameters are known, that is the mu and
this sigma is known, then using this one for a specific value of this x, if we do this
integration numerically, then we will get what is the probability from minus infinity to
that particular, to that specific value.
(Refer Slide Time: 11:19)
So, now every time the numerical integration instead of doing that one, there are some
standard tables are also available, that we will discuss in a minute. So, we can make use
of that table, but that table is generally for the standard normal distribution and we can
use that standard normal distribution for the calculation of this probability and that we
will discuss, while we are applying to a different problems.
So, the probability of a random variable, that is x being less than or equal to a specific
value x is equal to the area under the normal p d f curve bounded by x equals to that
specific value x and the probabilities also equal to the ordinate of the corresponding to
the x equals to this x in the normal c d f curve. So, what is mean is that. So, this is the
total area, that we are talking about from this integration, that what we are getting, if we
just refer to the, those curves that is so, this is.
So, up to any specific point, whether it is from any specific value if we come, so, it is the
total area up to that point below this curve and so, and that for that specific point,
whatever the area is obtained, that is the probability, which is indicated here by its
ordinate value. So, what we are referring here is that is that this area that is for say this is
your p d f. So, any specific value that we are talking about that x, so, up to this, whatever
the total area is covered here, so, this area for this value x, will be reflected in the c d f.
So, if this x corresponds to here, this x , now the c d f curve, that we are, that we have
obtained is so, this ordinates, so, this value, this value here, this height or this value is
nothing, but equal to this area.
(Refer Slide Time: 12:52)
So, this is what how you know that this p d f and c d f is related to, so, this integration
when we are talking about, this integration when are talking about, this integration is
basically referred to the area and that the curves. So, far as p d f is concerned and the
ordinate value in case of the c d f. So, now we will see that that a standard normal
distribution. So, you can you can now understand, that even though this integration, it
can be done numerically. Now, depending on the choice of this mu and sigma so, that
curve will change.
(Refer Slide Time: 13:42)
So, once - so, there can be some many means infinite numbers of the combinations so,
where the for a specific value of mu and the specific value of sigma square, so depending
on this mean and sigma square, that curve will change. So, every time you are supposed
to do that numerical integration, if you want to do that integration for the specific normal
distribution. So, instead of that, if we use the concept of this standard normal
distribution, then for the standard normal distribution, thus mean is 0 and the variance is
1, So, for that specific curve, if we can do this integration and if the value is available,
then for any such distribution, we can use the same table to know what is that integral
value or what is the probability, corresponding probability. So, this is what is explained
here, this standard normal distribution as you know, has the mean, mu equals to 0 and
variance sigma square is equals to 1. So, the standard normal table, lists the cumulative
probability values of the reduced variate z equals to x minus mu by sigma.
So, this is important that when, so, if this x is following a normal distribution, which is
having the parameter mean mu and variance sigma square, now you we have also seen in
this last module also in the function of random variable, you can refer to that. This is a 1
to 1 transformation and this is a linear transformation as well. So, if we do this type of
transformation, then this is also another random variable, which is denoted here z and is
generally known as their reduced variate, now this z will have, will also follow a normal
distribution, which will have your that mean is equals to 0 and the variance is equals to 1.
So, once we from any random variable, if it follows the normal distribution, now if we
do this type of transformation, that is the transformation is that x minus mu by sigma,
then the resulting random variable will have a, will also have the normal distribution, but
its parameter will be mean 0 and variance 1. Now, the table may be used conveniently
for calculating the probabilities associated with any normally distributed random variable
after normalizing it with the appropriate parameters. Now, with the appropriate
parameters, what is mean is that. So, with the mean of the original random variable, that
is x and divided it by the standard deviation of the original random variable x. So, before
you discuss further on this we will just see for the standard normal distribution how the
integral values or the standard statistical table is available to us.
(Refer Slide Time: 17:17)
This, we have we have discussed that if so, this is the bell shaped curve, this is the p d f
of this normal distribution, basically this is for the standard normal distribution and this
point is indicating that this is 0. So, which is the mean of this of this distribution and this,
for this standard normal distribution you know, how this forms look like. So, this is the
cumulative form that is shown here, which you can see which you know that that is your
f z, that is a reduced variate z and this is capital f; that means, we are referring to that the
cumulative distribution function for this standard normal distribution, which is equals to
1 by square root 2 pi, we are supposed to multiply it by sigma, but sigma you know that
this is 1. So, now, this is integration, integration from the left extreme, which is minus
infinity up to the specific value of that z and the integral form will be that e power half, I
can write half, then that this is that z minus 0 divided by sigma whole square. So, sigma
is 1 mu equals to 0. So, it will be only z square d z.
So, this is the form, what is the value is shown in this table. Now, for a specific value of
z this value refers to this area, if this parameter is mu equals to 0 and standard deviation
equals to 1. So, here that probabilities have shown here, you can see that this z values are
starting from this minus 1. So, this is zoomed here for. So, this is actually the entire
table, which is actually starting from minus 5 to plus 5. So, here some portion is zoomed
for the easy visibility. So, here you can see that, when the z is equals to 0. So, this
column will just read that 1, that is z equals to 0 and this row is for the second decimal.
So, this point corresponding to, so 0.5832 that you can see here, this is basically
corresponds to z equals to 0.21. So, this is the second decimal point. So, at z equals to
0.21 the value is 0.5832.
So, this is the value is given. Now, you know from the symmetry, if this is the standard
normal distribution. So, this is your 0 and as this is symmetric, you know that due to the
symmetry - so, what should be this area up to this 0 means, the total area under this curve
is 1. So, up to this 0, it would be 0.5, this is what is also shown in this curve, that is when
z equals to 0, the integral value is 0.5.
(Refer Slide Time: 20:40)
Anyway - so, we will now see a specific problem on this, how we can utilize this
property, is of this normal distribution for some specific problem in civil engineering.
So, the first problem, in a catchment, the total annual rainfall is estimated to be normally
distributed with a mean 150 centimeter and a standard deviation of 38 centimeter. Now,
here lies the question, that we just started with this class, that whenever we are starting
this in this lecture, whenever the problem we are starting with, we are stating that a
particular random variable is following a normal distribution. So, once we say the normal
distribution, that means, that parameters are also we have to know.
So, here the mean and standard deviation we know, that for the normal distribution there
are two parameters. So, those two parameters are also known, but the, what is
unanswered is that how to test that, here it is referred to that annual rainfall, how to know
that weather that is really a normal distribution or not. So, that question will be taken in
the subsequent lecture, how to test that its normality that we will see. But, here we are
starting the problem with the assumption that this follows a normal distribution with
some parameter value.
So, in a catchment the total annual rainfall is estimated to be normally distributed with a
mean of 150 centimeter and a standard deviation of 38 centimeter. So, what is the
probability that the annual rainfall in a certain year is between 100 centimeter and 170
centimeter. Second is what is the probability that the annual rainfall in a certain year is at
least 75 centimeter and the third is, what is the 75 percent dependable annual rainfall?
So, there are 3 questions is put and the 3 specific goal is kept here. The first question
reads that, what is the probability that, it will be between 100 centimeter to 170
centimeter. So, may be for the policy making in some cases, I need to know, that what is
the probability that it will be a normal year, normal year means, that the rainfall is near
normal and what is that that probability. So, if I say that, that probability is very high,
then we will be more confident to state. So, that particular year, the probability of this
much rainfall, that we get is having high probability and the second type of question,
what is thrown is that what is the probability that the annual rainfall in a certain year be
at least 75 centimeter.
This kind of issues comes, when we are interested about that minimum requirement of
the catchment. So, if this is. So, some stake holders for the water resource, there are a
different stake holders for example, the for the reservoir operation, we are supposed to
give the water for that multipurpose, one may be the power generation or for the
irrigation or for the industrial use. So, there are different users are there, whatever the
water is available, now the question is that to sustain the whatever the need to the
community, sometimes we need to ask the question, what is the probability that at least
this much rainfall or at least this much water is available to me. So, this is of that type of
question that at least 75 centimeters we can get from the thing.
And the last one is the what is the 75 percent dependable annual rainfall? That means,
that I want to be confident, I want to be confident that the with the 75 percent confidence
level, that I need to know that, yes, this much rainfall is certain at some probability level.
So, at 0.75 probability level, what is the value of this rainfall that I can expect. So, all
these answers will or based on the assumption that, at that point the rainfall is following
certain probability distributions. So, we are first developing, that probability thing, that
probability model and from that whatever the parameter that we are estimated, from that
we have to answer these two question in terms of a probabilistic manner.
So, we will take these problems now one after another. The first one is that 100
centimeter to 170 centimeter, now again if you go back, what is the information that is
supplied to us, the first is that it is, mean is that 150 centimeter and this variance, the
standard deviation is given 38 centimeter. So, you know that for this one, we do not have
any statistical table if of this of any normal distribution which is having mean 150 and
this standard deviation is 38. What we are having is that? We are having the table for this
0 mean and standard deviation equals to 1, means variance equals to 1 means, that
standard deviation is also 1.
So, first of all, as we are talking that, we have to do the transformation, we have to
generate, we have to obtain this reduce variate first, for which we are interested to know
the probability and after that we can make use of that table to give this type of answer.
So, first question we are taking, what is the probability that the annual rainfall in a
certain year is between 100 centimeter and 170 centimeter.
(Refer Slide Time: 26:30)
So, to find out this probability, what basically we are looking for, is the probability of
this random variable x. So, that annual rainfall, that we are getting in that catchment is,
we are the random variable is denoted by this x. So, this x will be between 100 to 170.
So, if we want to know this one. Then what we can say is that, so, if I just want to know
that this is that distribution form, so, obviously, if this is the origin I can say that, this is
0. So, the mean of this normal distribution, which is 0, which is away from this origin,
away from this 0. So, mean is shown to be that 150. So, this is your 150. So, what we are
interested to know that this random variable between 100 and 170. So, that this is your
170 and this is your 100.
.
(Refer Slide Time: 27:04)
So, what we are interested to know is this area. So, what we can, how to get this area,
what we cannot from the c d f point of view, if you want to know, that what is this area,
that is what we can get first, we can get the total area up to this 170, which will be
include this area as well minus this area, this up to 100. So, that difference will give you
this area. This is exactly what is written here also, that is probability x less than 1 less
than equals to 170 minus probability x less than 100.
Now, from here, so, for this type of normal distribution, that numerically integrated
values are not available to us, we are having the standard normal distribution only. So,
for that what we have to do, we have to first convert it to this standard normal
distribution and we have, we know that what should be the conversion to declare that the
new random variable will be having that 0 mean and standard deviation 1. So, that
conversion is nothing, but the x that specific value, that is 170 minus mean and mean is
150 here, divided by the standard deviation.
So, that is what, we are just mentioning that x minus mu divided by sigma. Now, x can
be of anything whatever is our interest and here mu, we have defined to be 150 and
sigma is your 38. Now, for this 170, the reduced variate will be, 170 minus 150 divided
by 38. Similarly, for the 100, this one the reduce variate will be 100 minus 150 divided
by 38. Basically, when we are getting these values, so, these values what basically what
we are doing it, we taking this full graph, we are basically converting it to another
normal distribution, for which the mean is at 0, standard deviation is 1 and we are here
instead of referring to 170, this 170 has converted to this value and this value is
somewhere here, which is showing to be this point.
And similarly, the 100 is converted to this one and this value is somewhere here, which
is a here. So, now, we are interested to know that this area, so, this area is equals to this
area to that original random variable scale. So, first of all, we have converted to like this
and because these values are available to us, then we will be referring to this, what the
standard distribution is available to us, then to get this answer that what is this area. So,
here is done this one, that is that we are reducing it from 175 minus 150 divided by 38
minus probability of z 100 minus 150 divided by 38. So, this corresponds to 0.53 and this
corresponds to minus 1.32.
Now, these values can also be directly obtained from the table, that we have shown just
now. So, first of all we will take this one, we will just discuss more on this minus 1.32.
Basically, we are using the symmetrical property. So, if the value is 0.53, then we have
to we have to find out what is this value is corresponding to.
(Refer Slide Time: 31:42)
Now, you see here. So, this is, so, 0.53, as we have telling that this column is the first
decimal of the reduce variate z and this row is the second one. So, we have to go to this
0.5 and here we have to go to that 0.03. So, this is your 0.53, the value for this 0.53,
which is 0.7019. So, this is what we have written that probability z less than equals to
0.53 is equals to 0.702.
Now, this 1 that z less than equals to minus 1.32 can also be obtained from the table and
can be directly used for, what is this value. Sometimes, what happens, in some text book
you will find that, using the symmetrical nature of this normal distribution table only one
side and preferably the positive side values are shown. So, what happens? In those cases
is done, that from this 0 it starts and go up to infinity means, here only 4 or 5 and those
values are given there.
So, once we know this value, that area then using this symmetrical property, it will be
easier to know that what is the area that we are looking for. For example, this point, for
this problem what we have seen is referring to minus 1.32. Now, we are interested to
know this area, now what is so, we are interested to this area. So, what we are doing is
that. So, up to this point, that is 0.53, we have, we came to know this from minus infinity
to this 0.53 that total area that we have seen is 0.7019 which is approximated to 0.702.
Now, what which area we have to, we have to deduct is this white area, that is from
minus infinity to point to 1.32. So, this much area we have to deduct. So, if this is the
negative side is not available to you, then using simply that positive side also, we can do
that analysis. So, how to do that one, so, we have to just replace this one. That is 1.32 to
this positive side, this area we have to know and this remaining this, whatever the white
area is there, which is equals to this side area. So, this is what is shown here that is
probability of z less than equals to minus 1.32 is equals to 1 minus probability of z less
than equals to 1.32. That is now we are using only the positive side of that value.
So, now that 1.32 value is also known from this table, that is this is z is 1.3 and second
decimal is 2, which is 0.9066. So, that 0.9066 is a 0.907. So, this we calculate. So, we are
getting the total area is 0.609. So, what is this total area, that we are getting here from
this reduced variate is 0.609, which is equal to this area as well. So, this is 0.609. So, we
can say that probability of the random variable x being between 100 to 170 is equals to
0.609. This is what is what is obtained here and we have explained that how we got this
corresponding values.
(Refer Slide Time: 35:39)
(Refer Slide Time: 35:46)
So, the table that we show here we should know here that the table that we show here is
continuing from this minus infinity to plus infinity, you can use, you can directly use this
table also to know that what is the probability for this minus 1.32, that also you can use
directly or if for some textbook only the positive side is given, using that the property of
this symmetry, then we can just convert it to this positive number, with the consideration
of the fact that the total area is 1. So, then you can use this, that particular table as well.
The second question is, what is the probability that the annual rainfall in the certain year
is at least 75 centimeter? So, before I start that one, so, the concept here is, that if you
understand that, how we are we are obtaining the probability for the normal distribution
after reducing it to this standard normal distribution, then all of this type of problem will
be straight forward, may be for the rest of the problem that we are discussing, we will
just discuss their inside, but how we are referring to this table and all we will not be
discussing in so details. So, I hope that you understood how we are getting the
probability by using the standard normal table.
So, here the second question is, that the probability, that the annual rainfall in the certain
year is at least 75 centimeters. So, what does it mean, at least 75 centimeters means, what
we are looking for is the probability x is greater than equals to 75. Now, this probability
x greater than equal to 75 can be, we are first converting it to the thing that 1 minus
probability of x less than equals to 75. So, why we are doing this 1, you know that now
the probability of x greater than equal to 75 plus probability of x less than equal to 75,
basically includes everything, all these all the area under the normal curve.
(Refer Slide Time: 38:06)
So, what we are talking here is that, if this normal distribution and if this is the mark for
this 75, then the probability x greater than equals to 75 plus probability x less than
equals to 75 is covering the entire area below this curve, that is which is 1. So, that is
why the standard normal distribution you know that is available for the less than equal to
values. So, as it is available for this less than equal to values, then that is why we are
interested to transfer it from this greater than equal to sign to this less than equals to sign.
Now, you know one thing is also important here, just to say you that, whether we will
include the greater than equal to sign will be included in the both the side, this question
may come to your mind, basically in the mathematical concept, what we should not, we
cannot include both the equality sign in both the side, either this should be equal to or
that should be equal to.
But, remember that, so far as the continuous random variable is concerned, this is of no
harm, because for the continuous random variable, if we want to know the probability of
a specific value, which is equal to the specific value is 0, that we discuss at the beginning
of this course. And this is what and this normal distribution you know, this is a
continuous random distribution, continuous probability distribution, So, for this one, if I
want to know, what is the probability that x equals to 75. That means, that probability x
equals to 75 is 0. So, what we can, what we can think in this way, that probability x
greater than equal to 75 is equals to 1 minus probability less than 75 plus 0 and that 0 is
nothing, but probability x equals to 75. So, whether you are including or excluding the
equality sign, it does not matter mathematically, so far as the distribution is continuous.
So, with this, now we will follow again the similar procedure, we will first we have to
transfer it to get that reduce variate, which is your 75 minus the mean 150 divided by the
standard deviation 38, which is equals to minus 1.97, this z less than equals to minus
1.97 can be directly obtained from the table or as I told, if only the positive side is
available, then also you can further transfer it to that 1 minus probability of z less than
equals to that plus 1.97. So, using this one, if you now again if you get that the value of
this z for 1.97 and you do this calculation basically, this probability less than equals to
this one will be equals to your 0.976.
So, the probability that the annual rainfall in a certain year is at least 75 centimeter is
your 0.976. So, for the catchment, that we are talking about, there is a high probability,
that 75 centimeter of rainfall. So, getting 75 centimeter of annual rainfall is highly
probable that is what we can infer.
(Refer Slide Time: 41:41)
And the third thing, that the last question that we have seen that let that 75 percent
dependable annual rainfall. Now, the 75 percent dependable annual rainfall, what we
mean is that, let that x 75, this is value is denoted that 75 percent dependable annual
rainfall. As I told that, this dependable rainfall means, I am assured that this much
rainfall can be obtained at this station with the probability level 0.75.
So, if that is the question then what I am looking from this one, so, if this is the
distribution of this annual rainfall, then I want to know a specific value for which, it is
located in at some point. So, this value I am looking for, what is this value, in such a way
that this is 75 percent dependable. This is 75 percent dependable means, whatever that
probability on the higher side of that value is equals to your 0.75.
Now, from the c d f point of view, if you want to know, that means, what is the value up
to which the total area covered under the p d f is equals to 0.25. This is what is the
concept and here now you can see that as we have denoted that x 75 is that particular
value, so that probability of x greater than equal to that value is equals to 0.75. So,
probability of z, when we are doing that transformation, when we are getting that
reduced variate; that means, that x 75 minus, that mean that 150 divided by the standard
deviation 38, that is the your reduced variate which is equals to 0.75.
The difference from this earlier 2 things, that we are discussed is that, there this
probability we are looking for and this specific value was known here it is reversed; that
means, I know the probability, I want to know what is the specific value both the things
can be done. Now, from that, so, if we just, so, this is the greater than equal to sign, if
you just refer to this one. That means, this area I am just talking about the 0.75.
Now, from the c d f we always know that this is the less than equals to, that means, if we
can convert into this side, then we can we can make use of that table. So, that is the
interest. So, for that what we are doing is that, instead of using the greater than equal to
sign, I am interested to know, what is the z less than equals to this quantity, this reduced
to variate, in that, what is happening to make this one this change, that we have to
subtract it from the total area below this curve. So, total area or the total probability is 1.
So, 1 minus, so, this is replaced by 1 minus probability z less than equals to this value,
which is equals to 0.75. Now, we are just doing this algebraic thing. So, this will be
equals to 0.25. So, this probability should be equals to 0.25.
Now, how we will make use of the table. So, now, how we will make the table is that, we
do not know what is z, what we know, we know what is the probability covered. So,
from the table, we have to find out the probability equals to 0.25 and corresponding to
that what is the z quintile that we can get. So that means, if we just go back to that
original table, so, instead of looking for this z first, we will look for this, which is the,
where is this probability is there. So, where is this probability there means, that for
example, here the probability is 0.8186. So, if I am interested for this probability, then I
should say that this is for the z value which is 0.91, similarly I have to look for the
probability which is equals to 0.25.
Now, this is; obviously, as this is the symmetrical you know that, from z is equals to 0.5
this is for the probability is equals to 0.5 the z value is 0, So, obviously, the z value will
be negative in that standard normal distribution to get the probability is equals to 0.25,
because the 0.25 is less than 0.5. Now, what we can do, if both the tables are available to
us like this, then we will, so, from the negative side, we will just find out where from this
0.25 is there and from there we will get what is this z value. If it is not available, then
what should I do, how can I use that only the positive side to know what is that quintile,
let us explain it first graphically.
(Refer Slide Time: 46:51)
That means, this is the standard normal distribution, this is your 0 and what is available
to us only this side, only the positive side from 0 onwards is available to us and what we
are interested to know is that, we are interested to know this value up to which the total
area is equals to 0.25. So, how we will get this one is that, we will find out for the value
where, so, if only the positive side is available, where this area is equals to 0.25. Why
0.25, because I know from this value, total area below this curve towards the positive
side is 0.5. Now, if I can say that this is equals to 0.25, that means, this area this, white
area is also 0.25. Now, suppose that I get this value equals to say z 1, at z 1 I have seen
that this from 0 to that z 1 value this probability is 0.25, that means, from z 1 to plus
infinity is the area is also 0.25.
Now, using the symmetry of this normal curve, what we can say that, this value will be
minus z 1. The value will be same only it will be on the other side of this 0. That means,
towards the negative side of this 0. So, to make this 1 equals to 0.25, this will be a minus
z 1. This can be done, in case from some textbook you get only the positive side values
of the normal distribution.
(Refer Slide Time: 48:44)
Now, here you can see that from this standard normal table, the value of the reduced
variate z corresponding to the non exceedence probability of 0.75 is 0.675. So, the value
of the z corresponding to the non exceedence probability of 0.25 is minus 0.675. So, the
value that the z 1 here, that we are referring to, is your, if you see the table, you will see
that it 0.675. So, that means, here that we are value, we are referring to it is minus 0.675.
If this is the thing, what we are doing is that. So, this is your now the reduced variate z,
what we have seen it earlier, that is x 75 minus 150 by 38, that this probability is equals
to 0.25, that means, this value which is nothing, but the z is equals to minus 0.675. This
is what is explained here. So, this is equals to minus 0.675 and by doing this 1 x 75 is
124.35, that means, the 75 percent dependable annual rainfall for that catchment is
124.35 centimeter.
(Refer Slide Time: 50:07)
We will go to the next problem, where also the similar the normal distribution is used
and this problem is on the testing that strength of concrete and you know that the
characteristics strength of the concrete is determined based on this 95 percent
dependability. Now, what is done you know that there are some several concrete cubes
of specific dimension 150 millimeter by 150 millimeter by 150 millimeter is casted and
they are cured in some identical condition and for 28 days at the end of 28 days, we
generally get the value of the crushing strength in the laboratory and different crushing
strength are obtained.
Now, from there, we have to see that, what is the strength we have to draw, because
obviously, if we go for the different concrete cube, many concrete cube for all the values.
So, as many concrete cube, those many values we will get. So, we have to conclude
something probabilistically. That type of problem is taken as the second problem here
that 24 concrete cubes are prepared and cured under the identical process the crushing
strengths of the concrete cube are then tested, the observed strength are shown in a table.
The question is, what is the mean and standard deviation of the strength of the cube and
what is the 95 percent dependable strength of the concrete cubes. Is it acceptable as M 20
grade concrete.
Now, you have to know that, this M 20 great concrete means, it is characteristics strength
is 20 Newton per millimeter square. So, this 24 cubes are tested for their crushing
strength and this values are shown here. So, what we are assuming that this follows a
normal distribution. We are first calculating their mean and their standard deviation. So,
first the mean is that 24.96, you know that will summation divided by 24 is giving that
24.9676 Newton per millimeter square. And second one is that the standard deviation,
the sample estimate you know, that 1 by n minus 1 x i minus x bar square from i equals
to 1 to n, here n equals to 24.
(Refer Slide Time: 51:50)
(Refer Slide Time: 52:03)
(Refer Slide Time: 52:54)
(Refer Slide Time: 53:24)
So, using this one we get the standard deviation is equals to 4.1594. So, we got these two
values now, the 96 percent dependable crushing strength is similar to the same that in the
earlier problem, we called that 75 percent dependable rainfall. So, that means, the
probability of x greater than equals to x 95, it is denoted as x 95, following the same
principle in the earlier problem. This minus that mean divided by the standard deviation
we are getting their reduce variate.
We are getting it the less than equal to sign, that is 1 minus probability of this and we are
getting that this equals to 0.05. Now, we have to find out the probability equals to 0.05
for which reduce variate. So, here you can see that, from the standard normal table, the
value of the z corresponding to 0.95 is 1.645. So, the value of z corresponding to 0.05 is
minus 1.645. So, this x 95 point that minus 24.9676 divided by their standard deviation.
This is the reduced variate, which should be equals to minus 1.645. So, x 95 is 18.125
from this experiment that we got. So, the x 95 that is 95 percent dependable that strength
is less than 20 Newton per millimeter square which is the requirement for the M 20 grade
concrete. So, we can conclude that it is not acceptable as M 20 concrete.
(Refer Slide Time: 54:23)
So, we can similarly we can go for the other problems, for the other civil engineering
aspects, for example, we have we can, there is another problem on this settlement of this
bridge pier, which is also important so far as the practical application is concerned. So,
we will take up that problem in the next class and that using. So, there also we will be
using the same concept and the same standard normal distribution table to get some
answers based on the probability models.
And, gradually also that, what we will take up that problem also. So, from that data, we
have to conclude that what distribution it is following through this probability plotting,
that we will take up and then also we will take up the central limit theorem, that we
started with in this class. So, we will continue again in this same topic in the next lecture.
Thank You.